Question

The magnitude of vectors $ \overrightarrow{A}$, $\overrightarrow{B}$ and $ \overrightarrow{C} $ are 3, 4 and 5 units respectively. If $ \overrightarrow{A} + \overrightarrow{B} = \overrightarrow{C}$, the angle between $ \overrightarrow{A} $ and $ \overrightarrow{B} $ is

• A. $\pi / 2$
• B. $ \cos^{ - 1} (0 . 6)$
• C. $ \tan^{ - 1} \, (7/5)$
• D. $ \pi / 4 $

Answer 1

The Correct Option is A

Let $ \theta $ angle between $ \overrightarrow{A} $ and $ \overrightarrow{B} $
Given : $ A = | \overrightarrow{A} | = 3 \, units$
$ B+ | \overrightarrow{B} | = 4\, units $
$ C = | \overrightarrow{C} | = 5 \,units $
$ \overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} $
$\therefore ( \overrightarrow{A} + \overrightarrow{B} ) \cdot ( \overrightarrow{A} + \overrightarrow{B} ) = \overrightarrow{C} \cdot \overrightarrow{C} $
$ \overrightarrow{A} \cdot \overrightarrow{A} + \overrightarrow{A} \cdot \overrightarrow{B} + \overrightarrow{B} \cdot \overrightarrow{A} + \overrightarrow{B} \cdot \overrightarrow{B} = \overrightarrow{C} \cdot \overrightarrow{C} $
$ A^2 + 2 A B \cos \theta + B^2 = C^2$
$9 + 2AB \cos \theta + 16 = 25$ or $ 2 A B \cos \theta = 0 $
or $ \cos \theta = 0 \therefore \theta = 90^\circ$.